problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 3.1.3. Suppose that the real numbers \(a, b, c > 1\) satisfy the condition
\[
\frac{1}{a^{2}-1}+\frac{1}{b^{2}-1}+\frac{1}{c^{2}-1}=1
\]
Prove that
\[
\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \leq 1
\] | Solution. Notice that if $a \geq b \geq c$ then we have
$$\frac{a-2}{a+1} \geq \frac{b-2}{b+1} \geq \frac{c-2}{c+1} \quad ; \quad \frac{a+2}{a-1} \leq \frac{b+2}{b-1} \leq \frac{c+2}{c-1}$$
Chebyshev's inequality affirms that
$$3\left(\sum_{c y c} \frac{a^{2}-4}{a^{2}-1}\right) \leq\left(\sum_{c y c} \frac{a-2}{a+1}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,503 |
Example 3.1.4. Let \(a, b, c, d, e\) be non-negative real numbers such that
\[
\frac{1}{4+a}+\frac{1}{4+b}+\frac{1}{4+c}+\frac{1}{4+d}+\frac{1}{4+e}=1.
\]
Prove that
\[
\frac{a}{4+a^{2}}+\frac{b}{4+b^{2}}+\frac{c}{4+c^{2}}+\frac{d}{4+d^{2}}+\frac{e}{4+e^{2}} \leq 1.
\] | Solution. The hypothesis implies that $\sum_{\text {cyc }} \frac{1-a}{4+a}=0$. We need to prove that
$$\sum_{c y c} \frac{1}{4+a} \geq \sum_{c y c} \frac{a}{4+a^{2}} \Leftrightarrow \sum_{c y c} \frac{1-a}{4+a} \cdot \frac{1}{4+a^{2}} \geq 0$$
Assume that $a \geq b \geq c \geq d \geq e$, then
$$\begin{array}{c}
\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,504 |
Example 3.1.5. Suppose that $a, b, c, d$ are four positive real numbers satisfying $a + b + c + d = 4$. Prove that
$$\frac{1}{11 + a^2} + \frac{1}{11 + b^2} + \frac{1}{11 + c^2} + \frac{1}{11 + d^2} \leq \frac{1}{3}$$ | Solution. Rewrite the inequality in the following form
$$\sum_{c y c}\left(\frac{1}{11+a^{2}}-\frac{1}{12}\right) \geq 0$$
or equivalently
$$\sum_{c y c}(1-a) \cdot \frac{a+1}{a^{2}+11} \geq 0$$
Notice that if $(a, b, c, d)$ is arranged in an increasing order then
$$\frac{a+1}{a^{2}+11} \geq \frac{b+1}{b^{2}+11} \geq ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,505 |
Example 3.2.1. Suppose \(a, b, c, d\) are positive real numbers such that
\[
a+b+c+d=a^{-1}+b^{-1}+c^{-1}+d^{-1}
\]
Prove the inequality
\[
2(a+b+c+d) \geq \sqrt{a^{2}+3}+\sqrt{b^{2}+3}+\sqrt{c^{2}+3}+\sqrt{d^{2}+3} .
\] | SOLUTION. A cursory look at this inequality will leave you hesitating. The relationship between the variables $a, b, c, d$ appears to be obscure and very hard to transform; moreover, the problem involves square roots. How can we handle this situation? Surprisingly enough, a simple way of applying Chebyshev can draw the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,507 |
Example 3.2.2. Suppose \(a, b, c\) are positive real numbers with sum 3. Prove that
$$\frac{1}{c^{2}+a+b}+\frac{1}{a^{2}+b+c}+\frac{1}{b^{2}+a+c} \leq 1$$ | Solution. The inequality is equivalent to
$$\sum_{c y c}\left(\frac{1}{c^{2}-c+3}-\frac{1}{3}\right) \geq 0 \Leftrightarrow \sum_{c y c}\left(\frac{a(a-1)}{a^{2}-a+3}\right) \geq 0$$
or
$$\sum_{c y c}\left(\frac{a-1}{a-1+\frac{3}{a}}\right) \geq 0$$
According to Chebyshev's inequality and the hypothesis that $a+b+c=3$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,508 |
Example 3.2.3. Let $a, b, c$ be positive real numbers and $0 \leq k \leq 2$. Prove that
$$\frac{a^{2}-b c}{b^{2}+c^{2}+k a^{2}}+\frac{b^{2}-c a}{c^{2}+a^{2}+k b^{2}}+\frac{c^{2}-a b}{a^{2}+b^{2}+k c^{2}} \geq 0$$ | Solution. Although this problem can be solved in the same way as example 2.1.1 is solved, we can use Chebyshev inequality to give a simpler solution. Notice that if \(a \geq b\) then for all positive real \(c\), we have \(\left(a^{2}-b c\right)(b+c) \geq\left(b^{2}-c a\right)(c+a)\), and
\[
\left(b^{2}+c^{2}+k a^{2}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,509 |
Example 1.1.6. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{i} \in[0, i]$ for all $i \in\{1,2, \ldots, n\}$. Prove that
$$2^{n} a_{1}\left(a_{1}+a_{2}\right) \ldots\left(a_{1}+a_{2}+\ldots+a_{n}\right) \geq(n+1) a_{1}^{2} a_{2}^{2} \ldots a_{n}^{2}$$ | Solution. According to AM-GM,
$$\begin{aligned}
a_{1}+a_{2}+\ldots+a_{k} & =1 \cdot\left(\frac{a_{1}}{1}\right)+2 \cdot\left(\frac{a_{2}}{2}\right)+\ldots+k \cdot\left(\frac{a_{k}}{k}\right) \\
& \geq \frac{k(k+1)}{2}\left(\frac{a_{1}}{1}\right)^{\frac{2}{k(k+1)}} \cdot\left(\frac{a_{2}}{2}\right)^{\frac{2}{k(k+1)}} \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,510 |
Example 3.2.4. Let \(a, b, c\) be positive real numbers. Prove that
\[
\sqrt{a^{2}+8 b c}+\sqrt{b^{2}+8 c a}+\sqrt{c^{2}+8 a b} \leq 3(a+b+c) .
\] | SOLUTION. Rewrite the inequality in the following form
$$\sum_{c y c}\left(3 a-\sqrt{a^{2}+8 b c}\right) \geq 0 \Leftrightarrow \sum_{c y c} \frac{a^{2}-b c}{3 a+\sqrt{a^{2}+8 b c}} \geq 0$$
or
$$\sum_{c y c} \frac{\left(a^{2}-b c\right)(b+c)}{(b+c)\left(3 a+\sqrt{a^{2}+8 b c}\right)} \geq 0$$
According to Chebyshev's... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,511 |
Example 3.2.5. Let \(a, b, c, d\) be positive real numbers such that \(a^{2}+b^{2}+c^{2}+d^{2}=4\).
Prove that
\[
\frac{1}{5-a}+\frac{1}{5-b}+\frac{1}{5-c}+\frac{1}{5-d} \leq 1
\] | Solution. The inequality is equivalent to
$$\begin{array}{l}
\sum_{c y c}\left(\frac{1}{5-a}-\frac{1}{4}\right) \leq 0 \Leftrightarrow \sum_{c y c} \frac{a-1}{5-a} \leq 0 \\
\Leftrightarrow \sum_{c y c} \frac{(a-1)(a+1)}{(5-a)(a+1)} \leq 0 \Leftrightarrow \sum_{c y c} \frac{a^{2}-1}{4 a-a^{2}+5} \leq 0 .
\end{array}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,512 |
Example 3.2.6. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers satisfying
$$a_{1}+a_{2}+\ldots+a_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}$$
Prove that the following inequality holds
$$\frac{1}{n^{2}+a_{1}^{2}-1}+\frac{1}{n^{2}+a_{2}^{2}-1}+\ldots+\frac{1}{n^{2}+a_{n}^{2}-1} \geq \frac{1}{a... | SOLUTION. WLOG, we may assume that $a_{1} \geq a_{2} \geq \ldots \geq a_{n}$. The hypothesis is equivalent to: $\square$
$$\frac{1-a_{1}^{2}}{a_{1}}+\frac{1-a_{2}^{2}}{a_{2}}+\ldots+\frac{1-a_{n}^{2}}{a_{n}}=0$$
Denote $S=\sum_{i=1}^{n} a_{i}$ and $k=n^{2}-1$. According to $\left({ }^{*}\right)$, the inequality can be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,513 |
Example 3.2.7. Suppose that \(a, b, c\) are positive real numbers with sum 3. Prove that
\[
\frac{1}{9-ab} + \frac{1}{9-bc} + \frac{1}{9-ca} \leq \frac{3}{8}
\] | SOLUTION. Let $x=b c, y=c a, z=a b$. The inequality becomes
$$\sum_{c y c} \frac{1}{9-x} \leq \frac{3}{8} \Leftrightarrow \sum_{c y c} \frac{1-x}{9-x} \geq 0$$
Suppose that $a_{x}, a_{y}, a_{z}$ are the coefficients we are looking for. We will rewrite the inequality to
$$\sum_{c y c} a_{x}(1-x) \cdot \frac{1}{a_{x}(9-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,514 |
Example 3.2.8. Let \(a, b, c\) be positive real numbers such that \(a^{4}+b^{4}+c^{4}=3\). Prove that
\[
\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca} \leq 1
\] | Solution. Let \(x=ab, y=ac\) and \(z=bc\). The inequality is equivalent to
\[
\begin{array}{c}
\frac{1-x}{4-x}+\frac{1-y}{4-y}+\frac{1-z}{4-z} \geq 0 \\
\Leftrightarrow \frac{1-x^{2}}{4+3 x-x^{2}}+\frac{1-y^{2}}{4+3 y-y^{2}}+\frac{1-z^{2}}{4+3 z-z^{2}} \geq 0
\end{array}
\]
Notice that \(a^{4}+b^{4}+c^{4}=3\) so \(x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,515 |
Example 3.2.9. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that
$$a_{1}+a_{2}+\ldots+a_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}$$
Prove the following inequality
$$\frac{1}{n-1+a_{1}^{2}}+\frac{1}{n-1+a_{2}^{2}}+\ldots+\frac{1}{n-1+a_{n}^{2}} \leq 1$$ | Solution. Rewrite the inequality to the following form:
$$\sum_{i=1}^{n}\left(\frac{1}{n-1+a_{i}^{2}}-\frac{1}{n}\right) \leq 0$$
or equivalently
$$\sum_{i=1}^{n} \frac{a_{i}^{2}-1}{n-1+a_{i}^{2}} \geq 0$$
Assume that $a_{1} \geq a_{2} \geq \ldots \geq a_{n}$. According to the hypothesis, we have
$$\sum_{i=1}^{n} \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,516 |
Example 4.1.1. Suppose that $x_{1}, x_{2}, \ldots, x_{n}$ are positive real numbers and $x_{1}, x_{2}, \ldots, x_{n} \geq 1$. Prove that
$$\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\ldots+\frac{1}{1+x_{n}} \leq \frac{n}{1+\sqrt[n]{x_{1} x_{2} \ldots x_{n}}}$$ | Solution. According to Lemma 2, it's enough to prove that
$$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}} \leq \frac{2}{1+a b} \forall a, b \geq 1$$
We can reduce this inequality to $(a-b)^{2}(1-a b) \leq 0$, which is obvious. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,519 |
Example 4.1.2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be real numbers lying in $(1 / 2,1]$. Prove that
$$\frac{a_{1} a_{2} \ldots a_{n}}{\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{n}} \geq \frac{\left(1-a_{1}\right)\left(1-a_{2}\right) \ldots\left(1-a_{n}\right)}{\left(n-a_{1}-a_{2}-\ldots-a_{n}\right)^{n}}$$ | Solution. The inequality is equivalent to
$$\sum_{i=1}^{n}\left(\ln a_{i}-\ln \left(1-a_{i}\right)\right) \geq n \ln \left(\sum_{i=1}^{n} a_{i}\right)-n \ln \left(n-\sum_{i=1}^{n} a_{i}\right)$$
Notice that the function $f(x)=\ln x-\ln (1-x)$, has the second derivative
$$f^{\prime \prime}(x)=\frac{-1}{x^{2}}+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,520 |
Example 1.1.7. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c} \leq \frac{1}{a b c}$$ | SOLUTION. Notice that $a^{3}+b^{3} \geq a b(a+b)$, so
$$\frac{a b c}{a^{3}+b^{3}+c^{3}} \leq \frac{a b c}{a b(a+b)+a b c}=\frac{c}{a+b+c} .$$
1.0. $A M$-GM inequality
21
Building up two similar inequalities and adding up all of them, we have the conclusion
$$\frac{a b c}{a^{3}+b^{3}+a b c}+\frac{a b c}{b^{3}+c^{3}+a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,521 |
Example 4.1.3. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geq 1 .
\] | SOLuTION. Although this problem has been solved using Hölder, a proof by Jensen's inequality is very nice, too. WLOG, we may assume that \(a+b+c=1\) normalize. Because \(f(x)=\frac{1}{\sqrt{x}}\) is a convex function, we obtain from Jensen's inequality that:
$$a \cdot f\left(a^{2}+8 b c\right)+b \cdot f\left(b^{2}+8 c ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,522 |
Example 4.1.4. Let \(a, b, c, d\) be positive numbers with sum 4. Prove that
\[
\frac{a}{b^{2}+b}+\frac{b}{c^{2}+c}+\frac{c}{d^{2}+d}+\frac{d}{a^{2}+a} \geq \frac{8}{(a+c)(b+d)}
\] | Solution. Denote \( f(x) = \frac{1}{x(x+1)} \), then \( f \) is a convex function if \( x > 0 \). According to Jensen inequality, we have
\[ \frac{a}{4} \cdot f(b) + \frac{b}{4} \cdot f(c) + \frac{c}{4} \cdot f(d) + \frac{d}{4} \cdot f(a) \geq f\left( \frac{ab + bc + cd + da}{4} \right) \]
which can be rewritten as
\[ ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,523 |
Example 4.1.5. Suppose that \(a, b, c\) are positive real numbers. Prove that
\[
\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3}{\sqrt{2}}.
\] | Solution. Notice that $f(x)=\sqrt{x}$ is a concave function. According to Jensen inequality, we have
$$\begin{aligned}
& \sum_{c y c} \sqrt{\frac{a}{a+b}}=\sum_{c y c} \frac{a+c}{2(a+b+c)} \cdot \sqrt{\frac{4 a(a+b+c)^{2}}{(a+b)(a+c)^{2}}} \\
\leq & \sqrt{\sum_{c y c} \frac{a+c}{2(a+b+c)} \cdot \frac{4 a(a+b+c)^{2}}{(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,524 |
Example 4.1.6. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{a}{\sqrt{4 b^{2}+b c+4 c^{2}}}+\frac{b}{\sqrt{4 c^{2}+c a+4 a^{2}}}+\frac{c}{\sqrt{4 a^{2}+a b+4 b^{2}}} \geq 1 .$$ | Solution. We may assume that $a+b+c=1$. Since $f(x)=\frac{1}{\sqrt{x}}$ is a convex function, according to Jensen's inequality, we have
$$a \cdot f\left(4 b^{2}+b c+4 c^{2}\right)+b \cdot f\left(4 c^{2}+c a+4 a^{2}\right)+c \cdot f\left(4 a^{2}+a b+4 b^{2}\right) \geq f(M)$$
where
$$M=a\left(4 b^{2}+b c+4 c^{2}\right)+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,525 |
Example 4.1.7. Let \(a, b, c\) be positive real numbers. Prove that
\[
\sqrt{\frac{a}{4a + 4b + c}} + \sqrt{\frac{b}{4b + 4c + a}} + \sqrt{\frac{c}{4c + 4a + b}} \leq 1
\] | SOLUTION. Notice that $f(x)=\sqrt{x}$ is a concave function, therefore by Jensen inequality we have
$$\begin{aligned}
\sum_{c y c} \sqrt{\frac{a}{4 a+4 b+c}} & =\sum_{c y c} \frac{(4 a+4 c+b)}{9(a+b+c)} \cdot \sqrt{\frac{81 a(a+b+c)^{2}}{(4 a+4 b+c)(4 b+4 c+a)^{2}}} \\
& \leq \sqrt{\sum_{c y c} \frac{(4 a+4 c+b)}{9(a+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,526 |
Example 4.1.8. Let \(a, b, c\) be positive real numbers such that \(a^{2}+b^{2}+c^{2}=3\). Prove that
\[
\sqrt{\frac{a}{a^{2}+b^{2}+1}}+\sqrt{\frac{b}{b^{2}+c^{2}+1}}+\sqrt{\frac{c}{c^{2}+a^{2}+1}} \leq \sqrt{3}
\] | Solution. Applying Jensen inequality for the concave function $f(x)=\sqrt{x}$, we have
$$\begin{aligned}
\sum_{c y c} \sqrt{\frac{a}{a^{2}+b^{2}+1}} & =\sum_{c y c} \frac{a^{2}+c^{2}+1}{3\left(a^{2}+b^{2}+c^{2}\right)} \cdot \sqrt{\frac{9 a\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\left(a^{2}+b^{2}+1\right)\left(a^{2}+c^{2}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,527 |
Example 4.2.1. Suppose that \(a, b, c\) are positive real numbers belonging to \([1,2]\). Prove that
\[a^{3}+b^{3}+c^{3} \leq 5 a b c\] | Solution. Let's first give an elementary solution to this simple problem. Since $a, b, c \in[1,2]$, if $a \geq b \geq c$ then
$$\begin{aligned}
a^{3}+2 \leq 5 a & \Leftrightarrow(a-2)\left(a^{2}+2 a-1\right) \leq 0 \\
5 a+b^{3} \leq 5 a b+1 & \Leftrightarrow(b-1)\left(b^{2}+b+1-5 a\right) \leq 0 \\
5 a b+c^{3} \leq 5 a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,528 |
Example 4.2.2. Suppose that \(a, b, c\) are positive real numbers belonging to \([1,2]\).
Prove that
\[
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 10
\] | Solution. The inequality can be rewritten as
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \leq 7$$
WLOG, we may assume that $a \geq b \geq c$, then
$$(a-b)(a-c) \geq 0 \Rightarrow\left\{\begin{array}{l}
\frac{a}{c}+1 \geq \frac{a}{b}+\frac{b}{c} \\
\frac{c}{a}+1 \geq \frac{c}{b}+\frac{b}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,529 |
Example 4.2.3. Let $x_{1}, x_{2}, \ldots, x_{2005}$ be real numbers belonging to $[-1,1]$. Find the minimum value for the following expression
$$P=x_{1} x_{2}+x_{2} x_{3}+\ldots+x_{2004} x_{2005}+x_{2005} x_{1}$$ | Solution. Because this inequality is cyclic, not symmetric, we cannot order variables as. If we rely on the relation $\left(x_{i}-1\right)\left(x_{i}+1\right) \leq 0$, we won't succeed either.
By intuition, we feel that the expression will attain its maximum if in the sequence $\left(x_{1}, x_{2}, \ldots, x_{2005}\rig... | -2003 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,530 |
Example 4.2.4. Given positive real numbers $x_{1}, x_{2}, \ldots, x_{n} \in[a, b]$, find the maximum value of
$$\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}-x_{3}\right)^{2}+\cdots+\left(x_{1}-x_{n}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\cdots+\left(x_{n-1}-x_{n}\right)^{2}$$ | Solution. Denote the above expression by $F$. Notice that $F$, represented as a function of $x_{1}$ (we have already fixed other variables), is equal to
$$f\left(x_{1}\right)=(n-1) x_{1}^{2}-2\left(\sum_{i=2}^{n} x_{i}\right) x_{1}+c$$
in which $c$ is a constant. Clearly, $f$ is a convex function $\left(f^{\prime \prim... | \max (F)=\left\{\begin{array}{l}
m^{2}(a-b)^{2} \text { if } n=2 m, m \in \mathbb{N} \\
m(m+1)(a-b)^{2} \text { if } n=2 m+1, m \in \mathbb{N}
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,531 |
Example 1.1.8. Prove that $x_{1} x_{2} \ldots x_{n} \geq(n-1)^{n}$ if $x_{1}, x_{2}, \ldots, x_{n}>0$ satisfy
$$\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\ldots+\frac{1}{1+x_{n}}=1 .$$ | Solution. The condition implies that
$$\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\ldots+\frac{1}{1+x_{n-1}}=\frac{x_{n}}{1+x_{n}}$$
Using AM-GM inequality for all terms on the left hand side, we obtain
$$\frac{x_{n}}{1+x_{n}} \geq \frac{n-1}{\sqrt[n-1]{\left(1+x_{1}\right)\left(1+x_{2}\right) \ldots\left(1+x_{n-1}\right)}}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,532 |
Example 4.2.5. Let $n \in \mathbb{N}$. Find the minimum value of the following expression
$$f(x)=|1+x|+|2+x|+\ldots+|n+x|, \quad(x \in \mathbb{R})$$ | Solution. Denote $I_{1}=[-1,+\infty), I_{n+1}=(-\infty,-n]$ and $I_{k}=[-k,-k+1]$ for each $k \in\{2,3, \ldots, n\}$. If $x \in I_{1}$ then $f(x)=\sum_{i=1}^{n}(1+x) \geq \sum_{i=1}^{n}(i-1)=\frac{n(n-1)}{2}=f(-1)$. If $x \in I_{n}$ then $f(x)=\sum_{i=1}^{n}(-1-x) \geq \sum_{i=1}^{n}(-i+n)=\frac{n(n-1)}{2}=f(-n)$.
Supp... | \left\{\begin{array}{l}
m(m+1) \text { if } n=2 m(m \in \mathbb{N}) \\
(m+1)^{2} \text { if } n=2 m+1(m \in \mathbb{N})
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,533 |
Example 4.2.6. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers belonging to $[0,2]$ such that $a_{1}+a_{2}+\ldots+a_{n}=n$. Find the maximum value of
$$S=a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}$$ | Solution. Applying the above lemma to the convex function $f(x)=x^{2}$, we get that $S$ attains the maximum if and only if $k$ numbers are equal to 2 and $n-k-1$ numbers are equal to 0. In this case, we have $S=4 k+(n-2 k)^{2}$. Because $a_{1}, a_{2}, \ldots, a_{n} \in[0,2]$, we must have $0 \leq n-2 k \leq 2$.
If $n=... | 2n+1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,534 |
Example 4.2.7. Suppose that \(a, b, c \in [0,2]\) and \(a+b+c=5\). Prove that
\[
a^{2}+b^{2}+c^{2} \leq 9
\] | Solution. Suppose that $a \leq b \leq c$. According to lemma 5, we deduce that $a^{2}+b^{2}+c^{2}$ attains the maximum if and only if $a=0$ or $b=c=2$. The first case $a=0$ is rejected because so, $4 \geq b+c=5$ is a contradiction. In the second case, we have $a=1$ and therefore $\max \left\{a^{2}+b^{2}+c^{2}\right\}=1... | 9 | Inequalities | proof | Yes | Yes | inequalities | false | 737,535 |
Example 4.2.8. Let $a_{1}, a_{2}, \ldots, a_{2007}$ be real numbers in $[-1,1]$ such that $a_{1}+a_{2}+\ldots+a_{2007}=0$. Prove that
$$a_{1}^{2}+a_{2}^{2}+\ldots+a_{2007}^{2} \leq 2006$$ | SOlUtion. Applying lemma 5, we deduce that the expression $a_{1}^{2}+a_{2}^{2}+\ldots+a_{2007}^{2}$ attains the maximum if and only if $k$ numbers equal 1 and $n-k-1$ numbers equal -1. $k$ must be 1003 and the last number must be 0, so we deduce that
$$a_{1}^{2}+a_{2}^{2}+\ldots+a_{2007}^{2} \leq 2006$$ | a_{1}^{2}+a_{2}^{2}+\ldots+a_{2007}^{2} \leq 2006 | Inequalities | proof | Yes | Yes | inequalities | false | 737,536 |
Example 4.2.9. Let \( x_{1}, x_{2}, \ldots, x_{n} \) be real numbers in the interval \([-1,1]\) such that \( x_{1}^{3} + x_{2}^{3} + \ldots + x_{n}^{3} = 0 \). Find the maximum value of \( x_{1} + x_{2} + \ldots + x_{n} \) | Solution. We denote \(a_{i}=x_{i}^{3}\) for all \(i \in\{1,2, \ldots, n\}\), then \(a_{1}+a_{2}+\ldots+a_{n}=0\). Notice that the function \(f(x)=\sqrt[3]{x}\) is concave if \(x \geq 0\) and convex if \(x \leq 0\). It's easy to get (as in lemma 5)
\[f(x)+f(y) \leq\left\{\begin{array}{l}
f(-1)+f(x+y+1) \text { if } x, y... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,537 |
Example 5.1.1 (Abel inequality). Let $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1} \geq y_{2} \geq \ldots \geq y_{n} \geq 0$ be real numbers. For each $k \in\{1,2, \ldots, n\}$, we denote $S_{k}=\sum_{i=1}^{k} x_{i}$. Suppose that $M=$ $\max \left\{S_{1}, S_{2}, \ldots, S_{n}\right\}$ and $m=\min \left\{S_{1}, S_{2}, \ldots... | Solution. Since both two parts of the inequality can be proved similarly, we only need to show the solution to the left inequality. Let $y_{n+1}=0$. By Abel's formula
$$\sum_{i=1}^{n} x_{i} y_{i}=\sum_{i=1}^{n}\left(y_{i}-y_{i+1}\right) S_{i} \geq \sum_{i=1}^{n} m\left(y_{i}-y_{i+1}\right)=m y_{1}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,538 |
Example 5.1.2. Let $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1} \geq b_{2} \geq \ldots \geq b_{n} \geq 0$ be positive real numbers such that $a_{1} a_{2} \ldots a_{k} \geq b_{1} b_{2} \ldots b_{k} \forall k \in\{1,2, \ldots, n\}$. Prove the following inequality
$$a_{1}+a_{2}+\ldots+a_{n} \geq b_{1}+b_{2}+\ldots+b_{n}$$ | Solution. By Abel formula, we deduce that
$$\begin{aligned}
\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} b_{i} & =\sum_{i=1}^{n} b_{i}\left(\frac{a_{i}}{b_{i}}-1\right) \\
& =\left(b_{1}-b_{2}\right)\left(\frac{a_{1}}{b_{1}}-1\right)+\left(b_{2}-b_{3}\right)\left(\frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}-2\right)+\ldots \\
& +\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,539 |
Example 5.1.3. Let \( x_{1}, x_{2}, \ldots, x_{n} \) be positive real numbers such that
\[ x_{1} + x_{2} + \ldots + x_{k} \geq \sqrt{k} \quad \forall k \in \{1, 2, \ldots, n\} \]
Prove the following inequality
\[ x_{1}^{2} + x_{2}^{2} + \ldots + x_{n}^{2} \geq \frac{1}{4} \left(1 + \frac{1}{2} + \frac{1}{3} + \ldots +... | SOLUTION. WLOG, assume that $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$. For each $k \in\{1,2, \ldots, n\}$, let $b_{k}=\frac{1}{\sqrt{k}}$. We will first prove that
$$2 \sum_{i=1}^{n} x_{i}^{2} \geq \sum_{i=1}^{n} x_{i} b_{i}$$
and then, $2 \sum_{i=1}^{n} x_{i} b_{i} \geq \sum_{i=1}^{n} b_{i}^{2}$
By Abel's formula, we ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,540 |
Example 5.1.4. Let $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ be real numbers such that
$$\begin{array}{l}
a_{1} \geq \frac{a_{1}+a_{2}}{2} \geq \ldots \geq \frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \\
b_{1} \geq \frac{b_{1}+b_{2}}{2} \geq \ldots \geq \frac{b_{1}+b_{2}+\ldots+b_{n}}{n}
\end{array}$$
Prov... | Solution. For each $k \in\{1,2, \ldots, n\}$, we denote $S_{k}=a_{1}+a_{2}+\ldots+a_{k}$ and $b_{n+1}=0$. By Abel's formula, we have
$$\sum_{i=1}^{n} a_{i} b_{i}=\sum_{i=1}^{n}\left(b_{i}-b_{i+1}\right) S_{i}=\sum_{i=1}^{n} i\left(b_{i}-b_{i+1}\right)\left(\frac{S_{i}}{i}\right)$$
According to Abel's formula again, we... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,541 |
Example 5.1.5. Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers such that $x_{1} \geq x_{2} \geq \ldots \geq x_{n} \geq$ $x_{n+1}=0$. Prove the following inequality
$$\sqrt{x_{1}+x_{2}+\ldots+x_{n}} \leq \sum_{i=1}^{n} \sqrt{i}\left(\sqrt{x_{i}}-\sqrt{x_{i+1}}\right)$$ | SOLUTION. Denote \( c_{i} = \sqrt{i} - \sqrt{i-1} \) and \( a_{i} = \sqrt{x_{i}} \). The inequality becomes
\[ \left(a_{1} c_{1} + a_{2} c_{2} + \ldots + a_{n} c_{n}\right)^{2} \geq a_{1}^{2} + a_{2}^{2} + \ldots + a_{n}^{2} \]
Suppose that \( b_{1}, b_{2}, \ldots, b_{n} \) are positive real numbers satisfying \( \sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,542 |
Example 1.1.9. Suppose that $x, y, z$ are positive real numbers and $x^{5}+y^{5}+z^{5}=3$.
Prove that
$$\frac{x^{4}}{y^{3}}+\frac{y^{4}}{z^{3}}+\frac{z^{4}}{x^{3}} \geq 3$$ | Solution. Notice that
$$\left(x^{5}+y^{5}+z^{5}\right)^{2}=x^{10}+2 x^{5} y^{5}+y^{10}+2 y^{5} z^{5}+z^{10}+2 z^{5} x^{5}=9$$
This form suggests the AM-GM inequality in the following form
$$10 \cdot \frac{x^{4}}{y^{3}}+6 x^{5} y^{5}+3 x^{10} \geq 19 x^{\frac{100}{19}}$$
Setting up similar cyclic results and adding up... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,543 |
Example 5.1.6. Let $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1} \leq b_{2} \leq \ldots \leq b_{n}$ be real numbers such that $a_{1}^{2}+a_{2}^{2}+\ldots+a_{k}^{2} \leq b_{1}^{2}+b_{2}^{2}+\ldots+b_{k}^{2} \forall k \in\{1,2, \ldots, n\}$. Prove that
$$a_{1}+a_{2}+\ldots+a_{n} \leq b_{1}+b_{2}+\ldots+b_{n}$$ | Solution. We prove this problem by induction. The case $n=1$ is obvious. Suppose that the problem has been proved for $n$ numbers already. We will prove it for $n+1$ numbers. Indeed, by Cauchy-Schwarz, we deduce that
$$\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n+1}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\ldots+b_{n+1}^{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,544 |
Example 5.1.7. Let $-1 < x_{1} < x_{2} < \ldots < x_{n} < 1$ and $y_{1} < y_{2} < \ldots < y_{n}$ be real numbers such that $x_{1} + x_{2} + \ldots + x_{n} = x_{1}^{13} + x_{2}^{13} + \ldots + x_{n}^{13}$. Prove that
$$x_{1}^{13} y_{1} + x_{2}^{13} y_{2} + \ldots + x_{n}^{13} y_{n} < x_{1} y_{1} + x_{2} y_{2} + \ldots ... | Solution. According to Abel formula, we note that
$$\begin{aligned}
\sum_{i=1}^{n} y_{i}\left(x_{i}^{13}-x_{i}\right)= & \left(y_{1}-y_{2}\right)\left(x_{1}^{13}-x_{1}\right)+\left(y_{2}-y_{3}\right)\left(x_{1}^{13}+x_{2}^{13}-x_{1}-x_{2}\right)+\ldots \\
& +\left(y_{n-1}-y_{n}\right)\left(\sum_{i=1}^{n-1} x_{i}^{13}-\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,545 |
Example 5.2.1. Let $a_{1}, a_{2}, . ., a_{n}$ be positive real numbers. Prove that
$$a_{1}+a_{2}+\ldots+a_{n} \geq n \sqrt[n]{a_{1} a_{2} \ldots a_{n}}$$ | Solution. WLOG, assume that $a_{1} a_{2} \ldots a_{n}=1$ (normalization). Let $a_{1}=\frac{x_{1}}{x_{2}}, a_{2}= \frac{x_{2}}{x_{3}}, \ldots, a_{n-1}=\frac{x_{n-1}}{x_{n}}, x_{1}, x_{2}, \ldots, x_{n}>0$, then $a_{n}=\frac{x_{n}}{x_{1}}$. The problem becomes
$$\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\ldots+\frac{x_{n-1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,546 |
Example 5.2.2. Suppose that \(a, b, c\) are the side-lengths of a triangle. Prove that
\[a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geq 0\] | Solution. Because $a, b, c$ are the side-lengths of a triangle, $a \geq b$ implies $a^{2}+b c \geq$ $b^{2}+c a$. By this property, we deduce that if $a \geq b \geq c$ then $a^{2}+b c \geq b^{2}+c a \geq c^{2}+a b$; also, $\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c}$. According to the Rearrangement inequality, we conc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,547 |
Example 5.2.3. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b} \geq a+b+c$$ | Solution. Applying Rearrangement inequality for the sequences $\left(a^{2}, b^{2}, c^{2}\right)$ and $\left(\frac{1}{b+c}, \frac{1}{\mathrm{c}+a}, \frac{1}{a+b}\right)$ (if $a \geq b \geq \mathrm{c}$, then these are both increasing), we get that
$$\sum_{c y c} \frac{a^{2}}{b+c} \geq \sum_{c y c} \frac{b^{2}}{b+c}$$
whi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,548 |
Example 5.2.4. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a+b}{a+c}+\frac{a+c}{b+c}+\frac{b+c}{a+b} \leq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}
\] | Solution. The inequality can be rewritten to
$$\sum_{c y c}\left(\frac{a}{b}-\frac{a}{b+c}\right) \geq \sum_{c y c} \frac{a}{a+c} \Leftrightarrow \sum_{c y c} \frac{a c}{b(b+c)} \geq \sum_{c y c} \frac{a}{a+c}$$
Consider the expressions $P=\sum_{c y c} \frac{a c}{b(b+c)}$ and $Q=\sum_{c y c} \frac{b c}{a(b+c)}$. By Re... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,549 |
Example 5.2.5. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \leq \frac{(a+b+c)^{2}}{a b+b c+c a}
\] | SOlutION. The inequality is equivalent to
$$\begin{aligned}
& \sum_{c y c} \frac{(a+b)(a(b+c)+b c)}{b+c} \leq(a+b+c)^{2} \\
\Leftrightarrow & \sum_{c y c} a(a+b)+\sum_{c y c} \frac{b c(a+b)}{b+c} \leq(a+b+c)^{2} \\
\Leftrightarrow & \sum_{c y c}\left(\frac{b c}{b+c}\right)(a+b) \leq a b+b c+c a
\end{aligned}$$
This la... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,550 |
Example 5.2.6. Let \(a, b, c, d\) be non-negative real numbers such that \(a+b+c+d=4\).
Prove that
\[
a^{2} b c + b^{2} c d + c^{2} d a + d^{2} a b \leq 4
\] | Solution. Suppose that $(x, y, z, t)$ is a permutation of $(a, b, c, d)$ such that $x \geq y \geq$ $z \geq t$, then $x y z \geq x y t \geq x z t \geq y z t$. By Rearrangement inequality, we deduce that
$$x \cdot x y z + y \cdot x y t + z \cdot x z t + t \cdot y z t \geq a^{2} b c + b^{2} c d + c^{2} d a + d^{2} a b$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,551 |
Example 5.2.7. Let \(a, b, c, d\) be positive real numbers. Prove that
\[
\left(\frac{a}{a+b+c}\right)^{2}+\left(\frac{b}{b+c+d}\right)^{2}+\left(\frac{c}{c+d+a}\right)^{2}+\left(\frac{d}{d+a+b}\right)^{2} \geq \frac{4}{9}
\] | Solution. WLOG, we may assume that $a+b+c+d=1$. Suppose that $(x, y, z, t)$ is a permutation of $(a, b, c, d)$ such that $x \geq y \geq z \geq t$, then
$$\frac{1}{x+y+z} \leq \frac{1}{x+y+t} \leq \frac{1}{x+z+t} \leq \frac{1}{y+z+t}$$
By Rearrangement inequality, we deduce that
$$\begin{aligned}
\sum_{c y c}\left(\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,552 |
Example 6.1.2. Let \(x, y, z, t\) be real numbers satisfying \(x y + y z + z t + t x = 1\). Find the minimum of the expression
\[5 x^{2} + 4 y^{2} + 5 z^{2} + t^{2}\] | Solution. We choose a positive number $l<5$ and apply AM-GM inequality
$$\begin{aligned}
l x^{2}+2 y^{2} & \geq 2 \sqrt{2 l} x y \\
2 y^{2}+l z^{2} & \geq 2 \sqrt{2 l} y z \\
(5-l) z^{2}+1 / 2 t^{2} & \geq \sqrt{2(5-l)} z t \\
1 / 2 t^{2}+(5-l) x^{2} & \geq \sqrt{2(5-l)} t x
\end{aligned}$$
Summing up these results, w... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,553 |
Let $a, b, c$ be real numbers satisfying $a+b+c=1$. By the AM - GM inequality, we have $a b+b c+c a \leq \frac{1}{3}$, therefore setting $a b+b c+c a=\frac{1-q^{2}}{3}(q \geq 0)$, we will find the maximum and minimum values of $a b c$ in terms of $q$.
If $q=0$, then $a=b=c=\frac{1}{3}$, therefore $a b c=\frac{1}{27}$. ... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,554 |
2.9 [Pham Huu Duc] For all positive real numbers $a, b$ and $c$,
$$\frac{1}{a^{2}+b c}+\frac{1}{b^{2}+c a}+\frac{1}{c^{2}+a b} \leq \frac{(a+b+c)^{2}}{3(a b+b c+c a)}\left(\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}\right) .$$ | Solution. Because the inequality is homogeneous, we may assume that \( p=1 \). Then \( q \in [0,1] \) and by the AM - GM and Schur's inequalities, we have \( \frac{(1-q^2)^2}{9} \geq 3r \geq \max \left\{0, \frac{1-4q^2}{9}\right\} \). After expanding, we can rewrite the given inequality as
\[
\begin{aligned}
f(r) = & -... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,556 |
2.10 [Nguyen Anh Tuan] Let $x, y, z$ be positive real numbers such that $x y+y z+z x+x y z=4$. Prove that
$$\frac{x+y+z}{x y+y z+z x} \leq 1+\frac{1}{48} \cdot\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right)$$ | Solution. Since $x, y, z>0$ and $x y+y z+z x+x y z=4$, there exist $a, b, c>0$ such that $x=\frac{2 a}{b+c}, y= \frac{2 b}{c+a}, z=\frac{2 c}{a+b}$. The inequality becomes
$$P(a, b, c)=\frac{(a+b+c)^{2} \sum_{\mathrm{cyc}}\left(a^{2}-b^{2}\right)^{2}}{(a+b)^{2}(b+c)^{2}(c+a)^{2}}-\frac{6 \sum_{\mathrm{cyc}} a(a+b)(a+c)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,557 |
2.11 [Nguyen Anh Tuan] For all nonnegative real numbers $a, b, c$
$$\sqrt{\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)}+\sqrt{\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right)}+\sqrt{\left(c^{2}-c a+a^{2}\right)\left(a^{2}-a b+b^{2}\right)} \geq a^{2}+b^{2}+c^{2}$$ | Solution. After squaring both sides, we can rewrite the inequality as
$$2 \sqrt{\prod_{\mathrm{cyc}}\left(a^{2}-a b+b^{2}\right)}\left(\sum_{\mathrm{cyc}} \sqrt{a^{2}-a b+b^{2}}\right) \geq\left(\sum_{\mathrm{cyc}} a b\right)\left(\sum_{\mathrm{cyc}} a^{2}\right)-\sum_{\mathrm{cyc}} a^{2} b^{2}$$
By the AM - GM inequa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,558 |
2.1 Let \(a, b, c\) be positive real numbers such that \(a+b+c=1\). Prove that
\[
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+48(ab+bc+ca) \geq 25
\] | Solution. We can easily check that $q \in[0,1]$, by using the theorem we have
$$L H S=\frac{1-q^{2}}{3 r}+16\left(1-q^{2}\right) \geq \frac{9(1+q)}{(1-q)(1+2 q)}+16\left(1-q^{2}\right)=\frac{2 q^{2}(4 q-1)^{2}}{(1-q)(1+2 q)}+25 \geq 25$$
The inequality is proved. Equality holds if and only if $a=b=c=\frac{1}{3}$ or $a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,560 |
2.2 [Vietnam 2002] Let $a, b, c$ be real numbers such that $a^{2}+b^{2}+c^{2}=9$. Prove that
$$2(a+b+c)-a b c \leq 10$$ | Solution. The condition can be rewritten as $p^{2}+2 q^{2}=9$. Using our theorem, we have
$$LHS=2 p-r \leq 2 p-\frac{(p+q)^{2}(p-2 q)}{27}=\frac{p\left(5 q^{2}+27\right)+2 q^{3}}{27}$$
We need to prove that
$$p\left(5 q^{2}+27\right) \leq 270-2 q^{3}$$
This follows from
$$\left(270-2 q^{3}\right)^{2} \geq p^{2}\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,561 |
2.3 [Vo Quoc Ba Can] For all positive real numbers $a, b, c$, we have
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}+11 \sqrt{\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}} \geq 17$$ | Solution. Because the inequality is homogeneous, without loss of generality, we may assume that \( p=1 \). Then \( q \in [0,1] \) and the inequality can be rewritten as
\[ \frac{1-q^{2}}{3 r} + 11 \sqrt{\frac{1-q^{2}}{1+2 q^{2}}} \geq 20 \]
Using our theorem, it suffices to prove
\[ 11 \sqrt{\frac{1-q^{2}}{1+2 q^{2}}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,562 |
2.4 [Vietnam TST 1996] Prove that for any $a, b, c \in \mathbb{R}$, the following inequality holds
$$(a+b)^{4}+(b+c)^{4}+(c+a)^{4} \geq \frac{4}{7}\left(a^{4}+b^{4}+c^{4}\right)$$ | Solution. If $p=0$ the inequality is trivial, so we will consider the case $p \neq 0$. Without loss of generality, we may assume $p=1$. The inequality becomes
$$3 q^{4}+4 q^{2}+10-108 r \geq 0$$
Using our theorem, we have
$$3 q^{4}+4 q^{2}+10-108 r \geq 3 q^{4}+4 q^{2}+10-4(1-q)^{2}(1+2 q)=q^{2}(q-4)^{2}+2 q^{4}+6 \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,563 |
2.6 [Darij Grinberg] If $a, b, c \geq 0$, then
$$a^{2}+b^{2}+c^{2}+2 a b c+1 \geq 2(a b+b c+c a)$$ | Solution. Rewrite the inequality as
$$6 r+3+4 q^{2}-p^{2} \geq 0$$
If $2 q \geq p$, it is trivial. If $p \geq 2 q$, using the theorem, it suffices to prove that
$$\frac{2(p-2 q)(p+q)^{2}}{9}+3+4 q^{2}-p^{2} \geq 0$$
or
$$(p-3)^{2}(2 p+3) \geq 2 q^{2}(2 q+3 p-18)$$
If $2 p \leq 9$, we have $2 q+3 p \leq 4 p \leq 18$, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,564 |
2.7 [Schur's inequality] For any nonnegative real numbers \(a, b, c\),
\[
a^{3}+b^{3}+c^{3}+3 a b c \geq a b(a+b)+b c(b+c)+c a(c+a)
\] | Solution. Because the inequality is homogeneous, we can assume that $a+b+c=1$. Then $q \in[0,1]$ and the inequality is equivalent to
$$27 r+4 q^{2}-1 \geq 0$$
If $q \geq \frac{1}{2}$, it is trivial. If $q \leq \frac{1}{2}$, by the theorem we need to prove that
$$(1+q)^{2}(1-2 q)+4 q^{2}-1 \geq 0$$
or
$$q^{2}(1-2 q) \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,565 |
2.8 [Pham Kim Hung] Find the greatest constant $k$ such that the following inequality holds for any positive real numbers $a, b, c$
$$\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+\frac{k(a b+b c+c a)}{(a+b+c)^{2}} \geq \frac{3}{8}+\frac{k}{3}$$ | Solution. For $a=b=1+\sqrt{3}$ and $c=1$, we obtain $k \leq \frac{9(3+2 \sqrt{3})}{8}=k_{0}$. We will prove that this is the desired value. Let $k_{0}$ be a constant satisfying the given inequality. Without loss of generality, assume that $p=1$. Then $q \in[0,1]$ and the inequality becomes
$$\frac{3\left(3 r+q^{2}\righ... | \frac{9(3+2 \sqrt{3})}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,566 |
Theorem 1 (Generalized Schur Inequality). Let $a, b, c, x, y, z$ be six non-negative real numbers such that the sequences $(a, b, c)$ and $(x, y, z)$ are monotone, then
$$x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b) \geq 0 .$$
Proof. WLOG, assume that $a \geq b \geq c$. Consider the following cases:
(i). $x \geq y \geq z$. The... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,567 |
Example 1.1.1. Let $a, b, c$ be three positive real numbers. Prove that
$$a+b+c \leq \frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b}$$ | SOLUTION. According to the identity
$$\frac{a^{2}+b c}{b+c}-a=\frac{(a-b)(a-c)}{b+c}$$
we can transform our inequality into the form
$$x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b) \geq 0$$
where
$$x=\frac{1}{b+c} ; y=\frac{1}{c+a} ; z=\frac{1}{a+b}$$
WLOG, assume that $a \geq b \geq c$, then clearly $x \leq y \leq z$. The conc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,568 |
Example 1.2.1. Let \(a, b, c, d\) be non-negative real numbers such that \(a+b+c+d=1\).
Prove that
\[a(a-b)(a-c)(a-d)+b(b-a)(b-c)(b-d)+c(c-a)(c-b)(c-d)+d(d-a)(d-b)(d-c) \geq \frac{-1}{432}\] | SOLUTION. We use the entirely mixing variable and the renewed derivative to solve this problem. Notice that our inequality is exactly
$$\frac{1}{432}(a+b+c+d)^{4}+\sum_{c y c} a(a-b)(a-c)(a-d) \geq 0$$
Notice that the inequality is clearly true if $d=0$, so we only need to prove that (after taking the global derivativ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,569 |
Example 1.17.3. Let $a, b, c, x, y, z$ be six real numbers in $\mathbb{I}$ satisfying
$$a+b+c=x+y+z, \max (a, b, c) \geq \max (x, y, z), \min (a, b, c) \leq \min (x, y, z)$$
then for every convex function $f$ on $\mathbb{I}$, we have
$$f(a)+f(b)+f(c) \geq f(x)+f(y)+f(z)$$ | SOlution. Assume that $x \geq y \geq z$. The assumption implies $(a, b, c)^{*} \gg (x, y, z)$ and the conclusion follows from Karamata inequality. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,570 |
Example 1.17.4. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers. Prove that
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right) \leq\left(1+\frac{a_{1}^{2}}{a_{2}}\right)\left(1+\frac{a_{2}^{2}}{a_{3}}\right) \ldots\left(1+\frac{a_{n}^{2}}{a_{1}}\right) .$$ | SOLUTION. Our inequality is equivalent to
$$\ln \left(1+a_{1}\right)+\ln \left(1+a_{2}\right)+\ldots+\ln \left(1+a_{n}\right) \leq \ln \left(1+\frac{a_{1}^{2}}{a_{2}}\right)+\ln \left(1+\frac{a_{2}^{2}}{a_{3}}\right)+\ldots+\ln \left(1+\frac{a_{n}^{2}}{a_{1}}\right)$$
Suppose that the number sequence $(b)=\left(b_{1},... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,571 |
Example 1.17.6. Suppose that $\left(a_{1}, a_{2}, \ldots, a_{2 n}\right)$ is a permutation of $\left(b_{1}, b_{2}, \ldots, b_{2 n}\right)$ which satisfies $b_{1} \geq b_{2} \geq \ldots \geq b_{2 n} \geq 0$. Prove that
$$\begin{array}{l}
\left(1+a_{1} a_{2}\right)\left(1+a_{3} a_{4}\right) \ldots\left(1+a_{2 n-1} a_{2 n... | SOLUTION. Denote $f(x)=\ln \left(1+e^{x}\right)$ and $x_{i}=\ln a_{i}, y_{i}=\ln b_{i}$. We need to prove that
$$\begin{array}{l}
f\left(x_{1}+x_{2}\right)+f\left(x_{3}+x_{4}\right)+\ldots+f\left(x_{2 n-1}+x_{2 n}\right) \\
\leq f\left(y_{1}+y_{2}\right)+f\left(y_{3}+y_{4}\right)+\ldots+f\left(y_{2 n-1}+y_{2 n}\right)
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,572 |
Example 1.17.7. Let $a, b, c, d$ be non-negative real numbers. Prove that
$$a^{4}+b^{4}+c^{4}+d^{4}+2 a b c d \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}+a^{2} c^{2}+b^{2} d^{2}$$ | SOLUTION. To prove this problem, we use the following lemma:
$\star$ For all real numbers $x, y, z, t$ then
$$2(|x|+|y|+|z|+|t|)+|x+y+z+t| \geq |x+y|+|y+z|+|z+t|+|t+x|+|x+z|+|y+t|$$
We will not give a detailed proof of this lemma now (because the next problem shows a nice generalization of this one, with a meticulous ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,573 |
Example 1.17.8. Let $a_{1}, a_{2}, \ldots, a_{n}$ be non-negative real numbers. Prove that
$$(n-1)\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}\right)+n \sqrt[n]{a_{1}^{2} a_{2}^{2} \ldots a_{n}^{2}} \geq\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2}$$ | SOLUTION. We realize that Turkevici's inequality is a particular case of this general problem (for $n=4$, it becomes Turkevici's). By using the same reasoning as in the preceding problem, we only need to prove that for all real numbers $x_{1}, x_{2}, \ldots, x_{n}$ then $\left(a^{*}\right) \gg\left(b^{*}\right)$ with
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,574 |
Example 1.17.9. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers with product 1. Prove that
$$a_{1}+a_{2}+\ldots+a_{n}+n(n-2) \geq(n-1)\left(\frac{1}{\sqrt[n-1]{a_{1}}}+\frac{1}{\sqrt[n-1]{a_{2}}}+\ldots+\frac{1}{\sqrt[n-1]{a_{n}}}\right)$$ | SOLUTION. The inequality can be rewritten in the form
$$\sum_{i=1}^{n} a_{i}+n(n-1) \sqrt[n]{\prod_{i=1}^{n} a_{i}} \geq(n-1) \sum_{i=1}^{n} \sqrt[n-1]{\prod_{j \neq i} a_{j}}$$
First we will prove the following result (that helps us prove the previous inequality immediately): if $x_{1}, x_{2}, \ldots, x_{n}$ are real... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,575 |
Example 1.17.10. Let $a_{1}, a_{2}, \ldots, a_{n}$ be non-negative real numbers. Prove that
$$(n-1)\left(a_{1}^{n}+a_{2}^{n}+\ldots+a_{n}^{n}\right)+n a_{1} a_{2} \ldots a_{n} \geq\left(a_{1}+a_{2}+\ldots+a_{n}\right)\left(a_{1}^{n-1}+a_{2}^{n-1}+\ldots+a_{n}^{n-1}\right)$$ | SOLUTION. We will prove first the following result for all real numbers $x_{1}, x_{2}, \ldots, x_{n}$
$$n(n-1) \sum_{i=1}^{n}\left|x_{i}\right|+n|S| \geq \sum_{i, j=1}^{n}\left|x_{i}+(n-1) x_{j}\right|$$
where $S=x_{1}+x_{2}+\ldots+x_{n}$. Indeed, let $z_{i}=\left|x_{i}\right| \forall i \in\{1,2, \ldots, n\}$ and $A=\{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,576 |
Example 1.17.11. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1} \geq a_{2} \geq \ldots \geq a_{n}$. Prove the following inequality
$$\frac{a_{1}+a_{2}}{2} \cdot \frac{a_{2}+a_{3}}{2} \cdots \frac{a_{n}+a_{1}}{2} \leq \frac{a_{1}+a_{2}+a_{3}}{3} \cdot \frac{a_{2}+a_{3}+a_{4}}{3} \cdots \frac... | Solution. By using Karamata's inequality for the concave function $f(x)=\ln x$, we only need to prove that the number sequence $\left(x^{*}\right)$ majorizes the number sequence $\left(y^{*}\right)$, where $(x)=\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ and $(y)=\left(y_{1}, y_{2}, \ldots, y_{n}\right)$, and for each $i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,577 |
Example 1.2.2. Let \(a, b, c, d\) be non-negative real numbers. Prove that
\[
a(a-b)(a-c)(a-d)+b(b-a)(b-c)(b-d)+c(c-a)(c-b)(c-d)+d(d-a)(d-b)(d-c)+a b c d \geq 0
\] | Solution. We use the global derivative as in the previous solution. Notice that this inequality is obvious due to Schur's inequality if one of the four numbers $a, b, c, d$ is equal to 0. By taking the global derivative of the left-hand side expression, we only need to prove that
$$\sum_{c y c}(a-b)(a-c)(a-d)+\sum_{c y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,578 |
Example 1.2.3. Let $a, b, c, d$ be non-negative real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$.
Prove that
$$\begin{array}{c}
a(a-b)(a-c)(a-d)+b(b-a)(b-c)(b-d)+c(c-a)(c-b)(c-d)+ \\
+d(d-a)(d-b)(d-c) \geq a b c d-1
\end{array}$$ | SOLUTION. We need to prove that
$$16 \sum a(a-b)(a-c)(a-d)+\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}-16 a b c d \geq 0$$
If $d=0$, the inequality is obvious due to the AM-GM inequality and Schur inequality (for three numbers). According to the mixing all variables method and the global derivative, it suffices to prove ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,579 |
Example 1.2.4. Let \(a, b, c, d, e\) be non-negative real numbers such that \(a+b+c+d+e=1\). Prove that
\[
\begin{array}{c}
a(a-b)(a-c)(a-d)(a-e)+b(b-a)(b-c)(b-d)(b-e)+c(c-a)(c-b)(c-d)(c-e)+ \\
+d(d-a)(d-b)(d-c)(d-e)+e(e-a)(e-b)(e-c)(e-d) \geq \frac{-1}{4320}
\end{array}
\] | SOLUTION. To prove this problem, we have to use two of the previous results. Our inequality is equivalent to
$$\frac{1}{4320}(a+b+c+d+e)^{5}+\sum_{c y c} a(a-b)(a-c)(a-d)(a-e) \geq 0$$
Taking the global derivative, we have to prove that
$$\frac{5}{860}(a+b+c+d+e)^{4}+\sum_{c y c}(a-b)(a-c)(a-d)(a-e) \geq 0$$
Due to t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,580 |
Example 1.2.5. Let $a, b, c, d, e$ be non-negative real numbers. Prove that
$$\begin{array}{c}
a(a-b)(a-c) \ldots(a-e)+b(b-a)(b-c) \ldots(b-e)+c(c-a)(c-b)(c-d)(c-e)+ \\
+d(d-a) \ldots(d-c)(d-e)+e(e-a)(e-b) \ldots(e-d)+a^{2} b c d+b^{2} c d e+c^{2} d e a+d^{2} e a b+e^{2} a b c \geq 0
\end{array}$$ | SOlutiOn. This problem is easier than the previous problem. Taking the global derivative for a first time, we obtain an obvious inequality
$$\sum_{c y c}(a-b)(a-c)(a-d)(a-e)+2 \sum_{c y c} a b c d+\sum_{c y c} a^{2}(b c+c d+d a) \geq 0$$
which is true when one of the numbers $a, b, c, d, e$ is equal to 0. Now we only n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,581 |
Example 1.2.6. Let $a_{1}, a_{2}, \ldots, a_{n}$ be non-negative real numbers such that $a_{1}+a_{2}+\ldots+a_{n}=$ 1. For $c=-9 \cdot 2^{2 n-7} n(n-1)(n-2)$, prove that
$$a_{1}\left(a_{1}-a_{2}\right) \ldots\left(a_{1}-a_{n}\right)+a_{2}\left(a_{2}-a_{3}\right) \ldots\left(a_{2}-a_{n}\right)+\ldots+a_{n}\left(a_{n}-a_... | To handle this problem, we need to prove it in the general case, that means, find an estimation of
$$F_{k, n}=\sum_{i=1}^{n}\left(a_{i}^{k} \prod_{j=1, j \neq i}^{n}\left(a_{i}-a_{j}\right)\right)$$
where the non-negative real numbers $a_{1}, a_{2}, \ldots, a_{n}$ have sum 1. After a process of guessing and checking in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,582 |
Example 1.3.1. Let $a, b, c, k$ be positive real numbers. Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+k}{b+k}+\frac{b+k}{c+k}+\frac{c+k}{a+k}$$ | SOLUTION. Notice that we can transform the expression $G(a, b, c)$ into
$$G(a, b, c)=\left(\frac{a}{b}+\frac{b}{a}-2\right)+\left(\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1\right)=\frac{(a-b)^{2}}{a b}+\frac{(a-c)(b-c)}{a c} .$$
WLOG, assume that $c=\min (a, b, c)$. Our inequality is equivalent to
$$\frac{(a-b)^{2}}{a b}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,583 |
Example 1.3.2. Let $a, b, c$ be positive real numbers. If $k \geq \max \left(a^{2}, b^{2}, c^{2}\right)$, prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a^{2}+k}{b^{2}+k}+\frac{b^{2}+k}{c^{2}+k}+\frac{c^{2}+k}{a^{2}+k}$$ | SOLUTION. Similarly as in the preceding inequality, this one is equivalent to
$$\frac{(a-b)^{2}}{a b}+\frac{(a-c)(b-c)}{a c} \geq \frac{(a-b)^{2}(a+b)^{2}}{\left(a^{2}+k\right)\left(b^{2}+k\right)}+\frac{(a-c)(b-c)(a+c)(b+c)}{\left(a^{2}+k\right)\left(c^{2}+k\right)}$$
WLOG, assume that $c=\min (a, b, c)$. It's suffic... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,584 |
Example 1.3.3. If $a, b, c$ are the side lengths of a triangle, then
$$4\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \geq 9+\frac{a^{2}+c^{2}}{c^{2}+b^{2}}+\frac{c^{2}+b^{2}}{b^{2}+a^{2}}+\frac{b^{2}+a^{2}}{a^{2}+c^{2}}$$ | SOLUTION. The inequality can be rewritten in the following form
$$\frac{4(a-b)^{2}}{a b}+\frac{4(c-a)(c-b)}{a c} \geq \frac{\left(a^{2}-b^{2}\right)^{2}}{\left(a^{2}+c^{2}\right)\left(c^{2}+b^{2}\right)}+\frac{\left(c^{2}-a^{2}\right)\left(c^{2}-b^{2}\right)}{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}$$
WLOG, w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,585 |
Example 1.3.4. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}+\frac{8(a b+b c+c a)}{a^{2}+b^{2}+c^{2}} \geq 11
\] | SOLUTION. Similarly, this inequality can be rewritten in the following form
$$(a-b)^{2}\left(\frac{(a+b)^{2}}{a^{2} b^{2}}-\frac{8}{a^{2}+b^{2}+c^{2}}\right)+(c-a)(c-b)\left(\frac{(a+c)(b+c)}{a^{2} c^{2}}-\frac{8}{a^{2}+b^{2}+c^{2}}\right) \geq 0$$
WLOG, assume that $c=\min \{a, b, c\}$. We have
$$\frac{(a+b)^{2}}{a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,586 |
Example 1.1.2. Let $a, b, c$ be positive real numbers with sum 3. Prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{3}{2 a^{2}+b c}+\frac{3}{2 b^{2}+a c}+\frac{3}{2 c^{2}+a b}$$ | SOLUTION. Rewrite the inequality into the following form:
$$\sum_{c y c}\left(\frac{1}{a}-\sum_{c y c} \frac{a+b+c}{2 a^{2}+b c}\right) \geq 0 \Leftrightarrow \sum_{c y c} \frac{(a-b)(a-c)}{2 a^{3}+a b c} \geq 0 .$$
Notice that if $a \geq b \geq c$ then
$$\frac{1}{2 a^{3}+a b c} \leq \frac{1}{2 b^{3}+a b c} \leq \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,587 |
Example 1.3.5. Let \(a, b, c\) be the side lengths of a triangle. Prove that
\[
\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\frac{c^{2}+a^{2}}{c^{2}+b^{2}}+\frac{b^{2}+c^{2}}{b^{2}+a^{2}} \geq \frac{a+b}{a+c}+\frac{a+c}{b+c}+\frac{b+c}{b+a} .
\] | SOLUTION. It is easy to rewrite the inequality in the following form
$$(a-b)^{2} M+(c-a)(c-b) N \geq 0$$
where
$$\begin{array}{l}
M=\frac{(a+b)^{2}}{\left(a^{2}+c^{2}\right)\left(b^{2}+c^{2}\right)}-\frac{1}{(a+c)(b+c)} \\
N=\frac{(a+c)(b+c)}{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}-\frac{1}{(a+c)(a+b)}
\end{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,588 |
Example 1.3.6. For all distinct real numbers $a, b, c$, prove that
$$\frac{(a-b)^{2}}{(b-c)^{2}}+\frac{(b-c)^{2}}{(c-a)^{2}}+\frac{(c-a)^{2}}{(a-b)^{2}} \geq 5$$ | SOLUTION. This inequality is directly deduced from the following identity
$$\frac{(a-b)^{2}}{(b-c)^{2}}+\frac{(b-c)^{2}}{(c-a)^{2}}+\frac{(c-a)^{2}}{(a-b)^{2}}=5+\left(1+\frac{a-b}{b-c}+\frac{b-c}{c-a}+\frac{c-a}{a-b}\right)^{2}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,589 |
Example 1.3.8. Let $a, b, c$ be distinct real numbers. Prove that
$$\frac{(a-b)^{2}}{(b-c)^{2}}+\frac{(b-c)^{2}}{(c-a)^{2}}+\frac{(c-a)^{2}}{(a-b)^{2}} \geq \frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$$ | SOLUTION. WLOG, we may assume that $c=\min (a, b, c)$. Taking into account the preceding example 1.3.1, we deduce that
$$G(a+b, b+c, c+a) \geq G(a+b-2 c, b+c-2 c, c+a-2 c)$$
Let now $x=a-c, y=b-c$, then it remains to prove that (after we consider $c=0$ )
$$\begin{array}{c}
\frac{(x-y)^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,591 |
Example 1.4.1. Let $a, b, c$ be non-negative real numbers. For each real number $k$, find the minimum of the following expression
$$S=\left(\frac{a}{b+c}\right)^{k}+\left(\frac{b}{c+a}\right)^{k}+\left(\frac{c}{a+b}\right)^{k}$$ | SOLUTION. Certainly, Nesbitt inequality is a particular case of this inequality for $k=1$. If $k \geq 1$ or $k \leq 0$ then it's easy to deduce that
$$\left(\frac{a}{b+c}\right)^{k}+\left(\frac{b}{c+a}\right)^{k}+\left(\frac{c}{a+b}\right)^{k} \geq \frac{3}{2^{k}}$$
If $k=\frac{1}{2}$, we obtain a familiar result as f... | \left(\frac{a}{b+c}\right)^{k} + \left(\frac{b}{a+c}\right)^{k} + \left(\frac{c}{a+b}\right)^{k} \geq \min \left\{\frac{3}{2^k}; 2\right\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,592 |
Example 1.4.2. Let \(a, b, c\) be the side lengths of a triangle. Prove that
\[
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3}{2} \leq \frac{2 a b}{c(a+b)}+\frac{2 b c}{a(b+c)}+\frac{2 c a}{b(c+a)}.
\]
or, in other words, prove that \(N(a, b, c) \leq 2 N\left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right)=2 N(a b, b ... | SOLUTION. First, we change the inequality to SOS form as follows
$$\sum_{cyc}\left(2 c^{2}-a b\right)(a+b)(a-b)^{2} \geq 0$$
WLOG, assume that $a \geq b \geq c$, then $S_{a} \geq S_{b} \geq S_{c}$. Therefore, it's enough to prove that
$$b^{2} S_{b}+c^{2} S_{c} \geq 0 \Leftrightarrow b^{2}\left(2 b^{2}-a c\right)(a+c)+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,593 |
Example 1.4.3. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{2 a b}{c(a+b)}+\frac{2 b c}{a(b+c)}+\frac{2 c a}{b(c+a)} \geq \frac{a+b}{2 c+a+b}+\frac{b+c}{2 a+b+c}+\frac{c+a}{2 b+c+a}
\]
or, in other words, prove that \(N\left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right) \geq N(a+b, b+c, c+a)\) | SOLUTION. Similarly to the preceding problem, after changing the inequality to SOS form, we only need to prove that
$$\frac{a\left(b^{3}+c^{3}\right)+b c\left(b^{2}+c^{2}\right)}{a b c \prod_{c y c}(a+b)} \geq \frac{2 a+3 b+3 c}{\prod_{c y c}(2 a+b+c)}$$
if $a, b, c \geq 0$ and $a \geq b \geq c$. Notice that $b^{3}+c^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,594 |
Example 1.5.2. Let \(a, b, c\) be the side lengths of a triangle. Prove that
\[ F(a, b, c) \leq 4 a^{2} b^{2} c^{2} F\left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right) \] | SOLUTION. Generally, the expression $F(a, b, c)$ can be represented in SOS form as
$$F(a, b, c)=\frac{1}{2} \sum_{cyc}(a+b-c)(a-b)^{2} .$$
Therefore we can change our inequality to the following
$$\sum_{cyc}\left(2 c^{2}\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)-(a+b-c)\right)(a-b)^{2} \geq 0$$
and therefore the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,595 |
Example 1.5.4. Prove that if $a, b, c$ are the side lengths of a triangle then
$$9 F(a, b, c) \geq 2 F(a-b, b-c, c-a)$$
and if $a, b, c$ are the side lengths of an acute triangle then
$$3 F(a, b, c) \geq F(a-b, b-c, c-a)$$ | SOLUTION. We will only prove the second part of this problem because the first part can be deduced similarly but simpler. Now suppose that $a, b, c$ are side lengths of an acute triangle. Clearly, if $x+y+z=0$ then
$$x^{3}+y^{3}+z^{3}+3 x y z-x y(y+x)-y z(y+z)-z x(z+x)=9 x y z .$$
Then, the inequality is equivalent to... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,597 |
Example 1.1.3. Let $a, b, c$ be the side lengths of a triangle. Prove that
$$\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}+\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}} \leq 3 .$$ | SOLUTION. By a simple observation, the inequality is equivalent to
$$\begin{array}{c}
\sum_{c y c}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right) \geq 0 \Leftrightarrow \sum_{c y c} \frac{\sqrt{a}+\sqrt{b}-\sqrt{c}-\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \geq 0 \\
\Leftrightarrow \sum_{c y c} \frac{\sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,598 |
Example 1.5.5. Let \(a, b, c\) be non-negative real numbers. Prove that
\[9 F\left(a^{2}, b^{2}, c^{2}\right) \geq 8 F\left((a-b)^{2},(b-c)^{2},(c-a)^{2}\right)\] | SOLution. We use the mixing all variables method, similarly as in the preceding problem. We can assume that \(a \geq b \geq c=0\). In this case, we obtain
\[
\begin{aligned}
F\left((a-b)^{2},(b-c)^{2},(c-a)^{2}\right) & =a^{6}+b^{6}+(a-b)^{6}+3 a^{2} b^{2}(a-b)^{2}-\left(a^{2}+b^{2}\right)(a-b)^{4} \\
& -a^{2} b^{2}\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,599 |
Example 1.6.1. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). Find the maximum of the following expressions
(a) \(S_{2}=2^{a b}+2^{b c}+2^{c a}\).
(b) \(\quad S_{4}=4^{a b}+4^{b c}+4^{c a}\). | SOLUTION. Don't hurry to conclude that $\max S_{2}=6$ and $\max S_{4}=12$ because the reality is different. We figure out a solution by the mixing variable method and solve a general problem that involves both $(a)$ and $(b)$. WLOG, assume that $a \geq b \geq c$ and $k \geq 1$ is a positive real constant. Consider the ... | k^{a b}+k^{b c}+k^{c a} \leq \max \left(3 k, k^{9 / 4}+2\right) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,600 |
Example 1.6.2. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). Find the minimum of the expression
\[3^{-a^{2}}+3^{-b^{2}}+3^{-c^{2}}\] | SOLUTION. We will again propose and solve the general problem: for each real number $k>0$, find the minimum of the following expression
$$P=k^{a^{2}}+k^{b^{2}}+k^{c^{2}}$$
Certainly, if $k \geq 1$ then $P \geq 3 k$ by AM-GM inequality. Therefore we only need to consider the remaining case $k \leq 1$. WLOG, assume that... | 1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,601 |
Example 1.6.3. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). Prove that
\[2^{4 a b}+2^{4 b c}+2^{4 c a}-2^{3 a b c} \leq 513 .\] | SOLUTION. In fact, this problem reminds us of Schur's inequality, only that we now have exponents (notice that if $a+b+c=3$ then Schur's inequality is equivalent to $4(a b+b c+c a) \leq 9+3 a b c$). According to example 1.6.1, we deduce that
$$2^{4 a b}+2^{4 b c}+2^{4 c a}=16^{a b}+16^{b c}+16^{c a} \leq 16^{9 / 4}+2=2... | 513 | Inequalities | proof | Yes | Yes | inequalities | false | 737,602 |
Example 1.7.1. Let $x, y, z$ be non-negative real numbers. Prove that
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+\frac{25(x y+y z+z x)}{(x+y+z)^{2}} \geq 8$$ | SOLUTION. WLOG, assume that $x \geq y \geq z$. Denote
$$f(x, y, z)=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+\frac{25(x y+y z+z x)}{(x+y+z)^{2}} .$$
We infer that
$$\begin{aligned}
f(x, y, z) & -f(x, y+z, 0)=\frac{y}{z+x}+\frac{z}{x+y}-\frac{y+z}{x}-\frac{25 y z}{(x+y+z)^{2}} \\
& =y z\left(\frac{25}{(x+y+z)^{2}}-\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,603 |
Example 1.7.2. Let \(x, y, z\) be non-negative real numbers. Prove that
\[
\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+\frac{16(xy+yz+zx)}{x^2+y^2+z^2} \geq 8
\] | SOLUTION. We use mixing variables to solve this problem. Denote \( x = \max \{x, y, z\} \) and
\[ f(x, y, z) = \frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} + \frac{16(xy + yz + zx)}{x^2 + y^2 + z^2}. \]
It is easy to see that the statement \( f(x, y, z) \geq f(x, y+z, 0) \) is equivalent to
\[ 16(x+y+z)^2 x(x+z)(x+y)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,604 |
Example 1.7.3. Let $a, b, c$ be non-negative real numbers with sum 1. Prove that
$$\frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}+\frac{27(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}} \geq 52$$ | SOLUTION. WLOG, assume that $a \geq b \geq c$. Denote
$$f(a, b, c)=\frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}+\frac{27(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}} .$$
We will prove that $f(a, b, c) \geq f\left(a, \sqrt{b^{2}+c^{2}}, 0\right)$. Indeed
$$\begin{array}{c}
f(a, b, c)-f\left(a, \sqrt{... | 52 | Inequalities | proof | Yes | Yes | inequalities | false | 737,605 |
Example 1.7.4. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}+\frac{24}{(a+b+c)^{2}} \geq \frac{8}{a b+b c+c a}$$ | SOLUTION. WLOG, assume that $a \geq b \geq c$. Denote
$$f(a, b, c)=\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}+\frac{24}{(a+b+c)^{2}}-\frac{8}{a b+b c+c a}$$
We get that
$$\begin{array}{l}
f(a, b, c)-f(a, b+c, 0) \\
=\frac{1}{(a+b)^{2}}-\frac{1}{(a+b+c)^{2}}+\frac{1}{(c+a)^{2}}-\frac{1}{a^{2}}-\frac{8}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,606 |
Example 1.7.5. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{4(a+b+c)(a b+b c+c a)}{a^{3}+b^{3}+c^{3}} \geq 5$$ | SOLUTION. WLOG, assume that $a \geq b \geq c$. Denote $t=b+c$ and
$$f(a, b, c)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{4(a+b+c)(a b+b c+c a)}{a^{3}+b^{3}+c^{3}}$$
We have
$$\begin{array}{l}
f(a, b, c)-f(a, b+c, 0) \\
=\frac{b}{c+a}+\frac{c}{a+b}-\frac{b+c}{a}+\frac{4(a+b+c)(a b+b c+c a)}{a^{3}+b^{3}+c^{3}}-\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,607 |
Example 1.7.6. Let $a, b, c$ be non-negative real numbers. Prove that
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+\frac{9 \sqrt{a b+b c+c a}}{a+b+c} \geq 6 .$$ | SOLUTION. WLOG, assume that $a \geq b \geq c$. We have
$$\begin{array}{c}
\sqrt{\frac{a b}{a+c}}+\sqrt{\frac{a c}{a+b}} \geq \sqrt{\frac{b \cdot b}{b+c}}+\sqrt{\frac{c \cdot c}{c+b}}=\sqrt{b+c} \\
\quad \Rightarrow \sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}} \geq \sqrt{\frac{b+c}{a}}
\end{array}$$
Let now $t=b+c$, then ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,608 |
Example 1.1.4. Let $x, y, z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove that
$$\frac{x^{2}+y z}{x \sqrt{2(y+z)}}+\frac{y^{2}+z x}{y \sqrt{2(z+x)}}+\frac{z^{2}+x y}{z \sqrt{2(x+y)}} \geq 1$$ | SOLUTION. We use the following simple transformation
$$\begin{aligned}
\sum_{c y c} \frac{x^{2}+y z}{x \sqrt{2(y+z)}} & =\sum_{c y c} \frac{(x-y)(x-z)+x(y+z)}{x \sqrt{2(y+z)}} \\
& =\sum_{c y c} \frac{(x-y)(x-z)}{x \sqrt{2(y+z)}}+\sum_{c y c} \sqrt{\frac{y+z}{2}} .
\end{aligned}$$
By the generalized Schur inequality, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,609 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.